If (p,q) is the vertex point of the graph of function f(x)=ax2+2ax+a+1, with f(a)=19, then p+2q+3a=....
Explanation
Given f(a)=19 then
f(x)=ax2+2ax+a+1
f(a)=19
a⋅a2+2a⋅a+a+1=19
a3+2a2+a−18=0
Factor using Horner's method to get
a3+2a2+a−18=0
(a−2)(a2+4a+9)=0
a−2=0∪(a2+4a+9)=0
a=2∪a2+4a+9=0
Thus we obtain a=2, and the quadratic function becomes
f(x)=ax2+2ax+a+1
f(x)=2x2+2⋅2x+2+1
f(x)=2x2+4x+3
Vertex point p
p=2a−b=2(2)−4=−1
q=f(p)=f(−1)
=2(−1)2+4(−1)+3
q=1
Therefore the value of p+2q+3a is
−1+2(1)+3(2)=−1+2+6=7